Strain

In an earlier blog, the one with the example of stretched rubber tubing, we included a description of the concept of strain.  Now, we try to make it more plausible, more precise, and to include shearing, as well as normal strain.

We'll start with the one-dimensional case again.  We choose a fixed coordinate system, and consider a surface somewhere inside a material body, at a point x = (x,y,z), where the plane tangent to the surface is the y-z plane.  That is, the normal vector n(x) = (1,0,0). 


Under the action of forces applied to the surface of the body, let the displacement of the point x in the x direction be u(x).  Next, consider a point at a distance a further along the x-axis,  at x+a.  What is the displacement u (x+a)?  We can find out by expanding u(x+a) in a Taylor series (Wikipedia has a good article on the latter).  The expansion is

 

where u' is the first derivative of u with respect to x, u" is the second derivative, and so on.  If we take a to be small enough, then we can drop all the terms with higher powers of a, keeping only the first two.  Then, for a sufficiently small,  we have [u(x+a) - u(x)]/a = du/dx.  This is called the normal strain in the x-direction.  (It's also simply the definition of the derivative of u(x)). Later we will use the observation that [u(x+a)-u(x)]/a  is equal to the fractional change in length of the line segment (originally of length a) joining the points x and x+a.


The material is said to be in tension when the normal strain is positive.  Note that the displacement is then larger at x+a than it is at x.  So as we noted with the rubber tubing, the displacement must increase in the direction of increasing x.  Furthermore, we noted that for materials that follow Hooke's law, the normal displacement is a linear function of x.  


In the previous blog, we found that for a body in mechanical equilibrium, the divergence of the stress tensor is zero, which means that the stress is constant (i.e., doesn't depend on x) for this one-dimensional example.  Therefore each point in the material along x must be in tension.  


For example, when you pull on the rubber tubing at one end (holding the left end fixed in space), the material at the point x moves to the right.  Therefore the material at the point x+a must also move to the right - otherwise the material at x+a must also move to the right, since it is under the same stress.  So the material at x+a must move the same distance to the right as the material at x, plus an additional distance in order to be in tension. For a material following Hooke's law, the fractional  (i.e., percentage) change in position of each point is constant.  


The purpose of all this discussion is to make Hooke's law seem plausible.  That is, if stress and strain are both constant in a one-dimensional linearly elastic material, it makes sense that they should be proportional to each other, with the coefficient of proportionality being a property of the material.  


As an aside, note that we are assuming the material is isotropic, i.e. that its properties are the same in all directions.  For example if the elasticity varied so that, for example, the rubber was "stiffer" at x+a than at x, it would be possible for the displacement at x+a to be less than at x.  


However, things get more complicated in the three-dimensional case, as we'll see in the next blog.  Since stress is a tensor, it seems reasonable that strain is also, and we need to find a linear relationship between these two tensors.

The divergence of the stress tensor

The law of conservation of momentum usually says that the net force on an object is equal to its rate of change of momentum.  But for a region inside the object, there has to be an additional term, to account for the possible flux of momentum through the enclosing surface.  So the correction to the last equation from the previous blog, for conservation of momentum, must be as shown below.  The arrows over P and n are to remind us that the momentum and unit normal are vectors, and similarly for the tilde over the stress tensor.  This equation states that the rate of change of momentum of material within the volume V is equal to the combined effects of 1) a source term, given by the force density f, and 2) a flux term, given by the stress tensor dotted into the unit vector normal to the surface.  

 This equation fits the general form a conservation law in continuum mechanics, with the net rate of change of a quantity  (mass, momentum, energy) being the sum of a source within the region being considered, and the flux of the quantity in through the surface.  We have a minus sign for the flux term, because we take n to be the outer normal vector.  I should add that "continuum mechanics" is the study of materials whose properties, like mass, momentum and energy, are continuous functions of their spatial coordinates, i.e., x, y and z.  One more comment:  The volume V and its enclosing surface S are fixed in space.  It's hard to imagine much of a change in momentum in a solid object being slowly deformed - by stretching, compressing, bending, shearing, etc.  Yet the force required to cause the deformation can be very large, so this equation must represent the small difference between two very large numbers.

In calculating the work done in deforming a body, we assume that the rate of change of momentum, dP/dt, is so small it can be neglected, and therefore we can take the terms on the right hand side of the above equation as being equal.  

We now continue the discussion of work.  To get started, we make use of a formula from vector calculus, known as the divergence theorem (you can read about it in Wikipedia).  It states that the surface integral above can be transformed into a volume integral, by taking the divergence of the stress tensor:

The integrands on both sides are vectors, because the stress is a tensor.  By combining this this result with the previous equation (and with the rate of change of momentum equal to zero), we find:

Since the volume of integration is arbitrary, the integrands must be equal everywhere, and this leads to the second and third equations above (this time omitting the arrow and tilde).  The vector force acting on an infinitesimal volume of material is equal to the gradient of the stress tensor within that same small volume.  For a material in equilibrium, whose properties are the same in all directions (i.e., is isotropic), the divergence of the stress tensor is zero.  

We know that an elastic material, according to Cauchy's definition (see the blog about stretching a piece of rubber tubing), is one in which the stress depends only on the strain.  So this means we should be able to find an expression for the strain, set equal to zero.  

Since the strain depends on the gradient of the deformation (see the blog just referred to), we expect to find a set of three second order differential equations (for i = 1,2,3) for the deformation vector u of an isotropic material.  These equations, plus boundary conditions, should have a unique solution.

Unfortunately, nearly all boundary value problems in elasticity are very difficult to solve analytically, so there aren't very many examples that can be used just to clarify the concepts.  

Next time, we'll talk about Hooke's Law, the one we used for the rubber tubing example, generalized to three dimensions.

The strain tensor, and the stress-strain relationship

In order to proceed from the simple example of tension in a rod to a more general formulation of stress and strain, we need to consider the work done in deforming an elastic material.

We start with the definition: work is equal to force times distance - for example, think of rolling a ball uphill.  You multiply the required force by the distance traveled to obtain the work done.  Note that the work is recoverable,  since you can let the ball roll back downhill.  In pushing the ball uphill, you are storing potential energy, which can be recovered as kinetic energy by letting it roll back down.  The stored energy will be completely recovered only if none of it is needed to overcome friction.  More generally, the process is reversible if none of the stored energy is converted into heat.

Similarly, the work required to deform an elastic body is the force acting at each point in the body times the displacement of that point from its position when no force was applied.  The deformation is reversible if the body returns to its original state when the force is removed.  Note that this force is the limit of the force per unit volume, as the volume shrinks to a point, and not the force per unit area.

This force was introduced in an earlier blog, "The stress principle of Euler and Cauchy," with the first equation shown below.  The stress vector was given in terms of the stress tensor in the blog "Cauchy's Stress Theorem,"  as shown in the second equation below.  This latter formula can be written in tensor notation as given in the third equation, with the convention that repeated subscripts are summed.  So when you see a repeated subscript, imagine a summation sign, with the sum going from k=1 to k=3.  In this case, the equation means that the ith component of the stress vector depends on all three components of the surface normal vector, and also on three components of the stress tensor.

Finally, substituting the right hand side of the third equation in place of the stress vector into the first equation will gives us the relationship between the force density vector and the stress tensor shown in the fourth equation.  Recall that the area S is the entire area enclosing the volume V, and the equation holds for any volume chosen within the material body.  Note that the integrand on the right is given by the ith component of a vector, which has (at most) three terms, given by the summation over the subscript k.  

The meaning of this last equation is the same as the first equation:  It states that the forces acting on the interior of a part of the material body may be entirely determined from the forces acting on the enclosing surface.

But wait a minute!  In the earlier blogs on the balance of forces, and on the stretched piece of rubber, we learned that the total force acting on any portion of a body in equilibrium is zero.  Since this is true for any volume, it follows that f =0 at each point in the material.  You can imagine a point in the body being pulled in various directions, and if the net force were not zero, the body would deform in the neighborhood of that point.  

The two sides of the last equation above are non-zero only if the material is in the process of being deformed. Just as with the ball on the hill, you're only working while you're rolling it, at least as far as the ball is concerned.  (The energy your body requires to hold it there is another story).  

So there is a term missing from the equation.  To find it, we need to consider the conservation of momentum, which we'll do in the next blog.

A simple example of strain - a stretched piece of rubber tubing

The practical goal of elasticity theory is often to calculate the deformation of an elastic body under an applied stress. We want to know how much force can be applied to the wing of an airplane before it breaks, or how much force it would take to crumple a car body in a collision, or how much weight a bridge can carry.

The situations I just described represent large deformations and the eventual failure, or rupture, of the object. But elasticity theory usually applies only to small deformations (say, 3% or less). Often, however, it only takes a small deformation to cause trouble. For calculations involving the strength of materials, one wants to know how much stress can be applied before the material yields, i.e. is no longer elastic.

An elastic material is one which has a "natural" state, that it tries to return to when an external force is applied, and will return to when that force is released. This statement applies to each point in the body. (A material that does not return to its initial state, but remains deformed, is said to be "plastic.")

Consider a point in an elastic body which has no applied stress - no force is being exerted on its surface. We think of this point as being attached to the material, so that if the body is deformed in some way by an applied stress, this "material point" may move, relative to a coordinate system that is fixed in space.

What we may want from the theory is a way to calculate the displacement, the distance and direction that any material point in the body moves due to an applied force. Alternatively, we may want to know the distribution of forces within the body, given a certain deformation.

To get started, imagine that you are holding an exercise band - or any length of rubber tubing with a handle at each end. Put one handle on the floor and hold it there with your foot, and use one hand to pull the other handle up away from the floor, as you might to exercise your bicep. We refer to the end of the band attached to the handle on the floor as "the foot end," and the other end as "the hand end."

Imagine a cross sectional area through the tubing, and the total force acting over that area on the part of tubing that is above that cross section - between it and the hand end. The total force, the integral of the stress vector over that area, must equal the force you applying with your hand, since you are hold the band very still.

Since this is true for any cross section through the band, it must be that the force of the material acting on itself (the integral of the stress tensor over the surface) is constant along the band. Because the rubber is uniform, it seems reasonable that the stress tensor is uniform over the cross section, and is therefore itself constant along the length of the band.

Suppose the band has the length l when no force is applied. Consider a one dimensional coordinate system, with the foot end of the band at x=0, and the hand end at x=l. Now when the band is stretched to some length l+dl, every point material point fixed in the band moves away from the foot end. The hand end moves the distance dl, but the foot end doesn't move at all.

Similarly, a material point in the band that was at the position x in the natural state will move to a new position x.' The displacement of the band due to the stretching is u=x'-x. We see that u(0)=0, and u(l) =dl.* Since we are considering a uniform, or homogeneous piece of rubber, whose elastic properties are the same throughout, it seems reasonable to expect the displacement to be a linear function of position. That is, u is a constant times x, and from the boundary conditions just given, u=(dl/l)x.

You can try this, by marking the band with a piece of chalk 1" from one end. Hold the band with both hands on a flat surface, alongside a tape measure, with the chalk mark near the left end. Suppose the band is 20 inches long. Hold the left end fixed at the zero end of the tape measure. Now move the right end 5" to the right, so that it has moved 25% of the initial ("natural") distance from the left end. You will see that the chalk mark has moved 1/4," which also represents a 25% stretch from its initial position.

It seems plausible that the fractional movement of each point along the band under an applied force will be the same, as long as the elastic properties of the band are uniform. On the other hand, if there were a weak region in the material, it would stretch more there.

We already know how to calculate the stress as a function of the applied force. In fact, in this simple case, the stress is constant, and equal and opposite to the applied force per unit area, from the discussion above. But how can we determine the displacement from the applied force, i.e., from the stress?

We see that the stress cannot be an explicit function of the displacement, since the displacement is not constant, but increases linearly from the foot end, along the band. However, the derivative of the displacement is constant, since du/dx=dl/l. This derivative is called the strain.

So we will postulate that for this piece of rubber, the stress is equal to the strain times a proportionality constant, which is a property of the material.  (This postulate is known as Hooke's Law, originally developed for a spring).  We write the equation below, where sigma is the normal stress on a surface cut through the band normal to x (the y-z plane), dl/l is the strain (which is dimensionless), and E is the Young's Modulus of elasticity. Note that E has the same units as the stress, i.e. force per unit area

Since the stress is constant and equal to the applied force per unit area, say P, this completes the solution to this simple problem. The strain is constant, and equal to the applied force per unit area, divided by the Young's Modulus. The strain is a constant fraction of the total length of the band. For rubber this fraction can be fairly large without causing permanent damage - 20 % or more. For other materials, like metals, the strain for elastic behavior must remain quite small.

From the equation above, the applied force must behave correspondingly - i.e., not exceed around 20% of the Young's Modulus for the rubber material used to make the exercise tubing. (Actually, you can get them with different wall thicknesses, and or elastic properties). In my case, I didn't have to pull very hard to stretch the band 5", probably with no more than 25 pounds of force. The tubing is 3/8" diameter, with a wall thickness of 1/16", so the cross-sectional area is a little more than 0.07 sq in. The force per unit area, which is the stress in the material, is therefore about 350 psi. So from the equation above, the elastic modulus for this particular piece of rubber is about 1750 psi.

According to the literature (Table 6-1, Barrett, Nix and Tetelman, The Principles of Engineering Materials, Prentice-Hall, 1973, P. 197), the elastic modulus for natural rubber varies from 1,000 to 10,000 psi.

Other materials have much larger moduli - Aluminum's, for example is 10 million psi. If I applied the same force to a piece of aluminum tubing with the same length and cross-section, its length would increase by 1750 x 10E-7, which is less than 0.2 mil, or 0.0002".

From the above discussion, we now have a basis for understanding Cauchy's definition of an elastic material: The stress is determined by the strain only, and that strain is not taken with respect to an arbitrary state but with respect to one particular state, the preferred or 'natural state of the material.'" (Truesdell, The Principles of Continuum Mechanics).

Since stress is a tensor, it seems likely that strain must be a tensor as well, so we'll leave that discussion for the next blog.

The balance of forces within a material under stress

The drawing on the left is intended to represent a 3-dimensional solid body, acted upon by force PR on the right, and PL on the left. Within the body is a surface S.

Imagine that we have a fixed coordinate system (x,y,z) such that the positive x axis is normal to the surface S. We use the label SR for S when it represents a part of the surface of the volume VR, and SL for the part of the surface of the volume VL. (VR and VL together give the total volume).

It seems clear that each component of PR and PL in the x, y and z directions must be equal in magnitude, but with opposite sign. Otherwise, since the net force on a body is equal to the rate of change of its momentum with time, the body would accelerate in the direction of the greater force.

In elasticity theory, it is usually assumed that the material body is somehow held fixed in space, i.e. with balanced applied forces.

Now let's consider the surface S within the body: Here S, SL and SR all occupy the same region in space, but SR has normal vectors pointing into the region VL, and SL's normal vectors point into VR. At each point on S, the normal vectors for SR and SL have components in x, y and z which are equal in magnitude and opposite in direction.

Let's start with SL. We choose a point (x,y,z) on SL, and take the normal vector there to point in the positive x direction (into VR). Then the normal vector has only one non-zero component, as shown in the Cauchy formula on the left.

As an aside, note that we have used the symmetry of the stress matrix, which allows us to equate the off-diagonal components as shown.

Using matrix multiplication, we find that each component of the stress vector contains only one component of the stress tensor, as shown below.

The stress vector is determined by only three components of the stress tensor - the normal stress in the 11 or xx direction, and two shear stresses in the yx and zx directions.

Since the normal vector is of unit length, i.e. n = (1,0,0), that is n1 = 1, the equations on the left show that the x component of the stress vector (that is, of the force acting on the surface SL) is given by the xx component of the stress tensor.

Similarly the y component of the force t is given by the yx component of the stress tensor, and the z component of force is given by the zx component of stress.

Put another way, the yx component of the stress tensor is the y component of the force acting on the x plane. By"x plane," we mean the plane tangent to the surface SL at the point (x,y,z) whose normal vector points in the x direction. It's actually the yz plane, but calling it the x plane makes it easier to identify with the yx component of the stress tensor.

Similarly, the zx component of the stress tensor is the z component of the force acting on the x plane.

We now have a simple example of a particular stress vector acting on a surface within a body. The next step is to consider the stress vector acting on the same plane tangent to the same surface, but now imagining that surface to to belong to the remaining part of the body.

Referring again to the drawing at the top, we use the same coordinate system we had earlier, and consider the surface SR, which is part of the surface of the volume VR. We again choose the same point (x,y,z) that we chose before, with the same plane tangent to SL (and therefore to SR), only this time the normal vector has components (-1,0,0) in our coordinate system.

All we have to do to get stress vector components acting at that point is to replace the x component of n with -1. We get the equations shown below, and find that the components have the same magnitude as before, only they are in the opposite (-x) direction. Since this is true for any point (x,y,z) on the surface drawn anywhere in the body, it must be that the the stress vectors are balanced. Note that the stress components depend only on the location of (x,y,z) and not on the orientation of the surface through this point.

Therefore the stress vectors are balanced. If we choose a different surface through the same point, the components of n will change (the coordinate system is the same), but the stress vectors will still balance.

Truesdell pointed out that this result is actually a consequence of the conservation of momentum: The sum of all the external forces is zero (that PL and PR are equal and opposite). Otherwise, if there is a net force, the body will accelerate, and will no longer be in equilibrium.

Instead, the body will deform as a function of time. The result will be that the surface S will change over time, as will the normal vector: n=n(t).

Furthermore, if the applied forces PL and PR are unequal, then the integrals of t over the surfaces SL and SR will be unequal, since each of those integrals must balance the applied forces on their respective regions VL and VR. From this it is clear that the stress vectors cannot balance, even at a given instant of time. From this we conclude that when the body accelerates, there must be a discontinuity in the stress tensor across the surface S. That is, the value of the stress components will be different if we approach S from the right (VR) or the left (VL) side. This is the subject of the theory of shock waves.

We won't pause now (maybe later) to study shock waves, but suffice it to say that we would expect the acceleration necessary to cause a jump condition within the body of a material would have to be quite large - say of the magnitude that occurs when a massive object from space accelerates towards the earth.

In the next blog, we will return to our main subject - reversible, elastic deformations of homogeneous materials - and introduce some of the concepts needed to actually solve problems.

The Stress Principle of Euler and Cauchy

Near the end of an earlier blog - the one called "vectors," I almost inadvertently mentioned something so important that Truesdell referred to it as "the defining principle of continuum mechanics." He called it The Stress Principle of Euler and Cauchy. Note that it is a basic assumption about the nature of continuous media, rather than something that can be derived.

This principle states that there is a set of stress vectors acting on the surface of any region in a material which completely represent the forces which are exerted on the region by the material outside.

We can put this statement in the form of the equation below. The left hand side (LHS) represents the volume integral of the i th component of the force density acting on each point inside the volume, and the right hand side (RHS) is the integral of the i th component of the stress vector acting on the surface of V.

The mass density of the material at each point is given by the Greek letter rho, the force density f is the force per unit mass, and the stress vector is the force per unit area acting on the surface of V.

Since the region of of interest in the material can be any shape or size, including arbitrarily small, it follows from this equation that we don't need to know the forces inside the region in order to calculate anything. We only need to know the forces on its surface.

The integral on the LHS represents the total force due to the exterior material which acts upon the interior material. Similarly, the interior material acts on the exterior, with the same total force, because we are considering bodies which are in equilibrium.

By equilibrium, we mean the following. When we first apply some combination of constant forces to a body, it moves - it changes its shape. When the movement stops, we say the body is in equilibrium. The vector sum of the external forces applied to the body are exactly balanced by the reactive forces of the body onto whatever is applying the external forces, and a similar statement applies to each region of the body we choose.

Note that this force balancing only applies to elastic materials, like rubber. For example, if you pulled hard enough on the ends of a length of aluminum rod, it would be permanently deformed. You could hold the rod without pulling on it at all, and it would still be deformed.

On the other hand, a long piece of aluminum (or copper, or steel) wire can be held horizontally at one end, and moved up and down vertically at the other end without any permanent deformation, as long as the distance traveled by the free end is small enough compared with the wire's length. So if you let go of the free end, the wire will return to its initial position.

This observation about the balance of interior and exterior forces also applies to the RHS of the equation, but it requires some further discussion, along with a drawing, so I'll save that for the next blog.

Cauchy formula in matrix form

The stress vector t is now written as a 3-component column vector. The equation tells how to write three equations for the components of t, by multiplying the stress matrix by the unit vector n.

The multiplication is done as follows (see equations below): the first component of t is given by multiplying the "11" component of sigma by the first component of n, etc.

Note how the first subscript of sigma corresponds to the t subscript, and the second subscript of sigma matches the n subscript.

Each component of the stress vector t depends on three components of the stress tensor, and on the three components of the surface normal vector n. Recall that t is the force applied at each point on the surface of a chosen region of the body by the surrounding material, and that n is the unit vector normal to that surface at the same point.

Put another way, here's the remarkable thing about Cauchy's theorem: it means that for every possible surface through a given point in the body - and their are an infinite number of such surfaces - all you need are the components of the stress tensor at that point. Then you can calculate the stress vector for any surface through that point, converting its normal vector into the stress vector with the Cauchy formula above.

The first, or x, component of t represents the force in the x-direction. If positive, the x component points out from the surface at that point; i.e., it is a tensile force. If negative, the x component points into the region enclosed by the surface, and is therefore compressive.

Similarly, the y and z components of t are in the y and z directions.

Note that the x component of t depends not only on the component of the normal vector n in the x direction, but also on the y and z components of n, and similarly for the y and z components of t.

The stress matrix element that corresponds to the x component of t and the x component of n is the -11- or -xx- element, and similarly for the y and z components. Later we will see why these components of the stress tensor are called normal stresses. The off-diagonal components are the shear stresses.

Stress tensor in matrix form

The 9 components of the stress tensor may be written in matrix form as shown above. In the lower of the two matrices above, the first number in the subscripts to the sigmas gives the row of the matrix, and the second number gives the column. So, for example the subscript "11" is used to represent row 1, column 1, and so on.

The subscripts also have another meaning: the three x, y, and z axes of our Cartesian coordinate system can be labelled 1, 2, and 3 respectively. Then the subscript "12" stands for "xy," and so on, allowing us to write the stress matrix as it's shown in the upper matrix.

An important simplification to elasticity theory is made possible by the fact that the stress tensor for most materials is symmetric: That is, the "xy" and "yx" components are equal, and similarly for "xz" and "zx," and for "yz" and "zy. Therefore the stress tensor has only six distinct components, or "unknowns" to be determined. They are the three diagonal components, and the three (different) components off-diagonal.

Vectors can also be written as matrices, with the three vector components as either a column or row of three numbers. I'll give an example in the next post, by writing the Cauchy formula in matrix form, followed by the use of matrix multiplication to write an equation for each component of the stress vector t.

Cauchy's stress theorem

Here we have Cauchy's theorem: The stress vector t can be written as the vector dot product of the stress tensor, sigma, and the unit vector normal to the surface, n. Whereas t depends on n, sigma does not. Sigma is a function only of position in the elastic body.

However, being a symmetric tensor (defined in the next blog), sigma also requires six numbers to specify - six of the nine components of the tensor.

So while a vector has only three components, a tensor has as many as nine (some of the components may be the same, as we'll see in the next blog).

This formula for Cauchy's theorem is in vector notation, a shorthand form which can be understood by writing it out in matrix form, which I'll add to the next post.

more on the stress vector, and the stress tensor

Here is the imaginary cube inside the rubber band, oriented so that the vertical faces are perpendicular to the pulling direction. The stress is perpendicular to the two vertical faces of the cube on the left and right sides, and parallel to all the other faces.

(I keep forgetting to mention, you can enlarge the drawing by pressing the ctrl and + keys, and then shrink back with crtl and -).

Imagine that the cube shrinks to an infinitesimal size. The normal and shear stresses will be associated with almost the same point in the material - and yet, they will be different because they represent the force on different surfaces, with different normal vectors n.

So we say that t is a function of both the position in the body (x), and the unit vector normal to the surface (n) under consideration. Thus the stress vector t is very different from most vectors in continuum mechanics, which are functions only of their position in space (and of time, in the case of time-dependent phenomena).

Putting it another way, it takes 6 numbers to specify the stress vector: 3 for the position x, and 3 for the surface with unit normal n, for which we want to calculate the force.

Thus each component of the stress vector (each of the x, y and z components) is a mathematical function of 6 variables. In order to develop a theory of stress, a new concept was invented which allowed an important simplification: a way of representing the stress vector with a new function, which depends only on the position x. This new idea has been called Cauchy's theorem,* and is stated in the beginning of the next post.

*According to Truesdell (The Classical Field Theories of Mechanics, by Clifford Truesdell and Richard Toupin, published in 1960 in Volume III of the Handbuch der Physik), it was the French mathematician Augustin-Louis Cauchy (1789–1857) who invented the concept of the stress tensor. As we get further into the theory, I think you will begin to appreciate the importance of this fundamental theorem, and the genius of the man who came up with it.

vectors

We all live in a three-dimensional space, in which objects can be located by their distances from an arbitrary point in the space - call it the "origin." To facilitate this, we imagine a rectilinear, orthogonal coordinate system. "Rectilinear," because the space can be defined with straight lines (rather than curves, as, say, on the surface of a sphere), and "orthogonal" because a point in the space can be uniquely determined by specifying its location on three mutually perpendicular directions.

In the drawing, we have marked off the three perpendicular axes, labelled x, y and z. The location of the tip of the arrow (or vector) in the drawing can be represented in standard notation by (a,b,c), where a is the distance from the origin along the x axis, and similarly for b and c along the y and z axes. We say that the vector x = (a,b,c).

In the previous blog I showed a drawing with a point x on the surface of an arbitrary volume within a solid body. Now I can say more precisely, that x is the vector from the origin of the coordinate system to the point on the surface, but to keep the drawing simpler I omitted the coordinate system. Anyhow, I wanted to explain this so you'd know why x is a vector.

Now, to continue that earlier discussion: We have the unit vector n perpendicular to the arbitrary surface drawn through the point located by the vector x. It's easier to talk about n if we choose a coordinate system so that the surface through x is perpendicular to one of the axes. For example, if this surface is perpendicular to the x axis, then n = (1,0,0). This notation means that n has a value of 1 in the x direction, and zero in the y and z directions. (Remember, this value 1 has whatever unit of length you choose, as long as you are consistent, and use the same unit for every length).

Now we can talk about the stress vector, t. What makes it so interesting to me is that t depends not only on the location in the solid defined by x, but also on the orientation of the surface we've chosen to examine, that contains the point defined by x.

Before we go any further, let's define stress: stress results from the force of a solid material acting on itself.

Imagine that we cut a rubber band, so that we have a single piece of material, and we pull on it from each end. We are then exerting a force on the ends (it has to be the same force on each end, for the material to hold still), but what about in the interior of the body?

Imagine a very small cube, somewhere inside the rubber, with its surfaces lined up so one pair is perpendicular to the direction we are pulling, and the others are therefore parallel to that direction. The forces acting on those surfaces are exactly balanced by forces in the the opposite directions, since we are holding the rubber band still - i.e., the elastic body is stationary. The material is actually pulling on itself.

From a microscopic point of view, we are stretching the bonds between molecules of rubber (or even stretching the long chain molecules themselves), and the magnitude of the stress depends on the strength of the molecular bonds. These molecular forces act only within short distances, so the force on each molecule is due only to other molecules nearby. Electrical shielding by the nearby molecules prevents long range effects, except in special cases - piezoelectric materials, for example - when long range forces are important.

But here we are confining ourselves to a continuum model, in which the stress and the properties of the material can be described by mathematically continuous functions in space. That's why we just say that the material "acts on itself." Note also that this means for any region of the elastic body we want to consider, the deformation of that region is entirely determined by the forces acting on its surface. Also, the sum of these forces, taking their directions into account, must add up to zero - otherwise the body would move.

The forces on the perpendicular surfaces mentioned above are perpendicular, or "normal" to the surfaces (they are equal and in opposite directions), whereas the forces on the parallel surfaces are parallel to them, or "shear," (and they are also equal and in opposite directions). Since we are pulling on the band, the normal forces are called "tensile." If we had a material we could squeeze, like a sponge, the normal forces would be "compressive."

I'll add a drawing of this cube next.

what is stress?

Consider an elastic body of some kind - any solid material will do. Anything from a sponge to a brick can be compressed or stretched if enough force is applied to it. Some materials are very strong in compression, but weak in tension (stretching), like a brick or piece of concrete. A rubber band or a rope might be strong in tension, but would be hard to compress just because of it's shape.

Anyway, the light blue object on the left represents an arbitrary part (volume) of this elastic body, which has to be imagined as three-dimensional, since I've only drawn it in two. This part must be inside the body, and could include some of its surface (but doesn't have to).

My drawing is a little off, but the point x is supposed to be on the imaginary surface of this arbitrary volume within the body (not necessarily on the surface of the body).

Now consider a unit vector n ("unit" because it has the value of one in whatever unit of length you want to use - inches, centimeters, etc) which is perpendicular to this imaginary surface right at the point x. By the way, I'm using bold face letters for vectors.

Maybe I should stop here and explain about vectors. A vector has both a value, say 1 cm, and a direction, like north or south. In physics, we use a coordinate system to specify both position in the space, and direction. Since we are talking about three-dimensional objects, we choose one point in the space as the origin, as a reference point. Then all other points in the space can be identified by their distance from the origin, and in what direction you would go from the origin to find them.

So the point x on the surface in the drawing can be represented by three numbers - x,y,z - which are the distances along each of the three perpendicular axes in a rectilinear coordinate system. (Later on we'll want to label the axes x1, x2 and x3 instead of x,y and z).

I'll close for now and try to make a drawing of a coordinate system, that might clarify the explanation somewhat.

I love tensors

At the request of a friend, I've recently begun to review some notes I prepared in the mid 1990's, when I was trying to understand some thin film cracking problems at the semiconductor company where I was employed. Although I've hardly thought about the subject since I retired (in 1997), looking at those notes immediately brought back my fascination with elasticity theory.

My love of tensors goes back to my graduate school days, and the study of special relativity and fluid mechanics. The texts that I really bonded with were in the incomparable theoretical physics series by Landau and Lifshitz, published in the 1950's and 60's, but still available, I think. I was equally inspired by the work of Clifford Truesdell in continuum mechanics, which I found in a handbook published in 1960, called the Handbuch der Physik (don't worry, he wrote in English).

Elasticity theory appeals to me because it requires the use of tensors to make sense out of it, and I would love to try to explain some of the concepts in a simple and straightforward way. That's what I plan to do with this blog.